Q. If the equation of a velocity profile over a plate is V = 2 y2/3, in which ‘v’ is the velocity in mis at a distance of ‘y’ metres above the plate, Determine the shear stress at y = 0 and y = 0.075m (or) 7.5 сm.

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Q. If the equation of a velocity profile over a plate is V = 2 y2/3, in which ‘v’ is the velocity in mis at a distance of ‘y’ metres above the plate, Determine the shear stress at y = 0 and y = 0.075m (or) 7.5 сm.
Given: µ = 0.835 N.S/m²

SOLUTION:
The veloicty profile over the plate is
V = 2y⅔
dV/dy = 2 * 2/3 * y – 1/3= (4/3) * y-⅓
Shear stress τ = μ(dV/dy)
μ =0.835 N. S / (m ^ 2)
At y = 0
τ = infty(infinite)
at y = 0.075m
τ = 0.835 * 4/3 * 1/((0.075)⅓)
Shear stress τ = 2.64N / (m ^ 2)

 

Q. A flat place of area 1.5 * 10 ^ 6 * mm² is pulled with a speed of 0.4 m/s relative plate located at a distance of 0.15 mm from it. From the force and power required to maintain this speed, if the fluid separating them is having viscosity as 1 poise.

SOLUTION:
Given:
Area of the plate, A = 1.5 * 10 ^ 6 * mm² = 1.5m² Speed of plate relative to another plate,
du = 0.4m / s
Distance between the plates,
dy = 0.15mm = 0.15 * 10-³ m
Viscosity 1 μ = 1 poise = 10 Ns/m²
Using equation we have
τ = μ* du/dy = 1/10 * 0.4/(0.15 * 10 -³)
= 266.66 N/m²
(i). Shear force, F = τ × area
= 266.66×1.5
= 400 N
(ii) Power* required to move the plate at the speed 0.4 m/sec.
= F×u = 400 * 0.4 =160 W
Note:
Power =F x u Nm/s = F x uW ( Nm/s =watt)

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